A Geometric Approach to Differential Forms by David Bachman

[mathwonk]

well i noticed he is at cambridge, but he still seemed to be claiming to rewrite the whole mathematical basis of physics so i figred he is most likely a nutcase anyway.

of course we could be wrong. i mean i acknowledge that i also am a pod person mascarading as a normal human being.

 

[dextercioby]

Is that David Hestenes,a guy who was at "Department of Physics and Astronomy,Arizona State University,Tempe,Arizona" ...?

I've got a lecture by him at the 1996 Fourth International Conference on Clifford Algebras and Their Applications to Mathematical Physics",Aachen,D called "Spinor Particle Mechanics".

Back then he didn't seem to be a crackpot.He's published in peer reviewed journals.I don't know what happened in between,there are 9 years,after all...

Daniel.

 

[Doodle Bob]

That is not however how I learned it, and why its somewhat confusing. For instance, typically in physics Clifford algebras primarily arise when you want to stick a spin geometry (read spinor bundles) on a manifold. This is topologically restricting from the getgo, amongst other things you need a choice of complex structure and I think the other is that the second STiefel Whitney class is identically zero.

I guess it just means I don't understand Geometric Algebra, b/c not only is their definition of differential forms/manifolds different than what I learned and how I use it daily, it also seems their 'Clifford algebras' are somewhat different than I learned. For instance one second of googling gives 4 camps
http://www.ajnpx.com/html/Clifford/4CliffordCamps.html

You all will have to forgive me for my lame answer to the posed question. I didn't realize that the questioner was asking the question from such a sophisticated point of view. I was not aware that anyone disagreed with how to define Clifford algebras, et al., having myself assumed that it was all decided already. Looking at my copy of Spin Geometry, I think that maybe the idea is to create spin-like structures on a broader class of manifolds besides those with zero 2nd Stiefel-Whitney classes.

I asked a math proffessor about this the other day, and he babbled something (he was clearly confused too) about how they are trying to generalize cross products and how their construction is really only good in dimensions 3 and 4.

Since any odd-diml. complex projective space (among other higher dimensional creatures) is spin, maybe he was referring to the Seiberg-Witten equations which use spin geometry (and hence Clifford algebras) but seem restricted to the 4-diml. case.


I really do find it hard to believe though, that differential forms will become completely obsolete. Technically, Riemannian geometry has replaced calculus, but you still need the basic 1-diml. real calculus to actually do anything.

 

[mathwonk]

here is my perspective on "new" algebras, as derived from the old fashioned education i received in the 60's and currently visible in books such as lang's algebra.

an (associative) "algebra" A (with identity), over a ring R, is an abelian additive group with an associative bilinear multiplication, for which an element called 1 acts as the identity, equipped with a ring map from R to A, "preserving identities".


Given any module M over R there is a universal such object T(M) called the tensor algebra of M over R. There is always a module map from M into T(M), and the image generates T(M) as an algebra.

If M is free of rank s over R, then T(M) is a non commutative polynomial ring over R generated by s "variables", which can be chosen to be any s free generators of M as a module.

The beauty of this object is, it contains in its DNA the data of all possible such algebras over R. I.e. if B is any associate R algebra with identit, equipped with a module map M-->B whose image generates B over R, then there is a unique surjective R algebra map T(M)-->B such that the composition M-->T(M)-->B equals the given map M-->B.

Hence the "new" algebra B, is merely a quotient T(M)/I, of the universal algebra T(M) by some ideal I. In this sense there are no new algbras of this type, as they are all constructed out of T(M).


For example, if S(M) is the "symmetric algebra" of M over R, which just equals the usual commutative polynomial algebra over R, with algebra generators or "variables" equal to the module generators of M, then S(M) = T(M)/I where I is the 2 - sided ideal generated by elements of form
x(tens)y - y(tens)x.

and if E(M) is the exterior algebra of M over R, (whose elements are linear combinations of wedge products of things like dx, dy, dz, when dx, dy, dz are generators of M over R), then E(M) is just the quotient of T(M) by the ideal generated by elements which contain repeated factors like x(tens)x.


Now the usual definition of a Clifford algebra is that it is an associative algebra with identity, built on a vector space M over a field R, plus a quadratic form q ("inner product"), as follows: the algebra C(M) is equipped with a module map M-->C(M) such that the image of the element x, in C(M) has square equal to q(x).1. I.e. if x is in M, and q(x) is its "squared length" under the form q, then in C(M), we have x^2 = q(x).1. And the elements of M generate C(M) as an algebra over R. morover C(M) is universal for all such algebras, i.e. every other one is a quotient of C(M).

But in particuilar C(M) is an associative algebra generated by M. Hence there is a unique surjective R algebra map T(M)-->C(M) realizing C(M) as a quotient of form T(M)/I for some unique ideal I in T(M), containing elements of form
x(tens)x - q(x).1, and presumably generated by these.

Now I fully admit to being a novice here, but i fail to see how anyone can fail to deduce from this that the key construction to understand in all of this is the tensor product.

Moreover, as the Clifford algebra involves extra structure which is not always present, namely the form q, it is clearly a more special derivative of T(M) than is the exterior algebra E(M), i.e. differential forms.

Furthermore, what "new" algebras are possible? unless they are non associative. (and mathematicians have also studied non associative algebras but i have not myself.)

Anyone claiming to construct a new associative algebra generated by elements of a module M, makes one wonder if they are unaware of the basic universal constructions that have been on the scene and even dominated it since the 1950's.

Of course this all concerns only the local, i.e. pointwise side of the story. The usefuleness of these constructs to physicists should be influenced, perhaps decidely, by their global manifestations in physics.


Notice that even if I am completely wrong, I have purposely given you enough data to decide for yourself.

If someone in a competing camp wishes to share more sophisticated and newer definitions for these concepts, I assume we will all be grateful.

Oh yes, and Riemannian geometry cannot possibly replace calculus, as Riemannian geometry also invovles an inner product which is unnecessary for intrinsic ideas of calculus.

 

[mathwonk]

OK, I have looked on the webpage http://modelingnts.la.asu.edu/GC_R&D.html and in particular perused the short simplified version of GA, intended for high school teachers. there is of course nothing there which is new in the mathematical sense, but some which would seem new to high school students (although I had some of this material in second grade! from a student teacher experimenting on us with trigonometry), an mr hestenes' goal there is to advocate incorporating some well known ideas of vector algebra, exterior algebra, and quadratic forms, into high school geometry, which he calls geometric algebra.

so he is not a real crackpot since he advocates something both useful and correct, and which he also seems to understand; but he is sort of a missionary, and hence comes on like a crackpot by advertising his mission in overly glowing terms, claiming he is going to revolutionize physics education and provide the universal answer to all communication problems between the two sciences, and harking back to the golden days of the 19th century, and so on.

this makes his non technical stuff sound a little fishy. but there is a similar movement by people who dress up like patch adams and try to sell calculus to reluctant students with books called "streetwise calc for dummies" and so on, and they are real mathematicians who have done some people some good, or at least some of my friends think so.

so i for one am glad mr hestenes is out there pumping for more use of vector algebra in high school and college. and although this stuff did get published in am. j. physics it seems, it would be hard for me to believe it occurs in any research math journals. but i have a finite amount of evergy and interest to devote to this type of thing. but i say in this case, more power to him.

i try to do exactly the same type of thing in my teaching, i.e. take known ideas, which are however not having the impact they should have at lower levels, and force them in there, hopefully after having understood them myself that is. i do it right here on this forum all the time. i am not talking about anything mathematical here that is not extremely well known to most practicing mathematicians. my very modest contribution to things like the discussion of clifford algebgras is just to pick up a book not everyone may have access to, read it quickly as a mathematician, and report back here to the best of my ability.

 

[Doodle Bob]

Oh yes, and Riemannian geometry cannot possibly replace calculus, as Riemannian geometry also invovles an inner product which is unnecessary for intrinsic ideas of calculus.

As I'm sure you know, standard calculus on the real numbers uses the Euclidean norm to define convergence of limits of sequences (among others), which can be derived from the Euclidean inner product, although I suppose one could develop most if not all of standard calculus by defining any old Hausdorff topology on R and defining convergence of sequences from there. Derivatives might end up looking a little strange, if the topology is...

Please take no offense. I was just being cheeky.

 

[mathwonk]

are you under the impression that all norms arise from inner products?

i.e. that all banach spaces are hilbert spaces?

 

[Doodle Bob]

are you under the impression that all norms arise from inner products?

i.e. that all banach spaces are hilbert spaces?

Of course not. I just know that the standard distance norm on R can be seen (if one wants to) as coming from the (albeit rather trivial) inner product on R.

 

[quantumdude]

Back to business!

OK school's out, and I'm back for real. Let's finish this book by the end of the summer! Hoo-rah!

I'd like to pick up from where we left off in the book: Exercise 3.18. I posted a solution, to which AKG commented. I haven't looked at his comments in a while, but I do have questions on them. Naturally anyone is free to answer.

Here's my solution to the Exercise.

Exercise 3.18
Let \omega=w_1dx \wedge dy +w_2dy \wedge dz +w_3dz \wedge dx.
Let A=<a_1,a_2,a_3> and B=<b_1,b_2,b_3> be vectors in \mathbb{R}^3.
Let C=[c_1,c_2,c_3] be a vector in T_p\mathbb{R}^3 such that C=k_1A+k_2B. So the set {A,B,C} are dependent. That implies that det|C A B|=0.

Explicitly:

<br /> det [C A B]=\left |\begin{array}{ccc}c_1&amp;c_2&amp;c_3\\a_1&amp;a_2&amp;a_3\\b_1&amp;b_2&amp;b_3\end{array}\right|<br />

<br /> det [C A B]=c_1(a_2b_3-a_3b_2)-c_2(a_1b_3-a_3b_1)+c_3(a_1b_2-a_2b_1)<br />

Now let \omega act on A and B. We obtain the following:

<br /> \omega (A,B)=w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)<br />

Upon comparing the expressions for det [C A B] and \omega (A,B) we find that \omega (A,B)=0 if w_1=c_3, w_2=c_1, and w_3=c_2. So the line l is the line that is parallel to the vector [w_2,w_3,w_1]. So I can write down parametric equations for l as follows:

<br /> x=x_0+w_2t<br />
<br /> y=y_0+w_3t<br />
<br /> z=z_0+w_1t<br />
[/color]

AKG responded thusly.

So anyways, you have it right, and if you want to choose a line parallel to P, you may want to throw in that extra bit that allows you to claim that P is in the plane.

I did not use your P though. I used a vector C that is in the plane spanned by A and B. I did that for the purpose of choosing a line l that is parallel to C, so that the plane spanned by A and B is guaranteed to contain l. The only thing I did not determine was the point (x_0,y_0,z_0), but this can be found easily knowing the vector parallel to l and the equation of the plane.

One more remark: You have A and B in R³, and C in the tangent space. It seems as though you should have them all in R³, or all in the tangent space.

That is not consistent with any of the reading thus far. The rest of the chapter discussed forms defined on T_p\mathbb{R}^n that act on vectors in \mathbb{R}^n. Am I misunderstanding something?

 

[mathwonk]

i tried to answer this exercise in post 91, or so, as follows:

"Also in post #81, Tom asked about solving ex 3.18. What about something like this?
a 1 form measures the (scaled) length of the projection of a vector onto a line, and a 2 form measures the (scaled) area of the projection of a parallelogram onto a plane. Hence any plane containing the normal vector to that plane will project to a line in that plane. hence any parallelogram lying in such a plane will project to have area zero in that plane.

e.g. dx^dy should vanish on any pair of vectors spanning a plane containing the z axis.'

does that make any sense?

i have forgotten now but it sems the point is that 2 forms on R^3 are decomposable? reducible? whatever?

 

[quantumdude]

i tried to answer this exercise in post 91, or so, as follows:

(snip)

does that make any sense?

Yes, it made sense. It's just that the next few exercise deal with the line that was to be found in 3.18, which is why I wanted an algebraic result. I'll chew on your answer a little longer and see if I can't answer the other questions with it.

 

[pmb_phy]

are you under the impression that all norms arise from inner products?

Are you under the impression that the norm doesn't arise from a inner products?

From the context presented here the norm of a vector is the square root of the inner product of a vector. The defnition of norm of of continuous when ... sorry but [f(x) = f(x)

 

[Rev Prez]

My next question is for the students:

Would any of you like to show this? Check my notes for how to show linearity and non-linearity (think superposition and scaling).
[/color]

Without going through all the steps, scaling returns |\omega\wedge\nu(cV_1,V_2)| = |c||\omega\wedge\nu(V_1,V_2)| \not= c|\omega\wedge\nu(V_1,V_2)| for c < 0.

Rev Prez

 

[mathwonk]

pmb phy, the point of my post was that calculus for normed spaces depends only on the norm, hence makes sense in any banach space. and yes, i am under the impression, as are most people, that there exist banach spaces which are not hilbert spaces. i.e. there are norms which do not arise from dot products. (sup norm on continuous functions on [0,1].)

the message is that the derivative is a more basic concept than is the dot product, since the derivative makes pefect sense, with exactly the same definition, in situations where the dot product does not. of course people may disagree, but to me the evidence seems clear.

 

[quantumdude]

OK, back to my quandry. I feel like I'm on the cusp of finally moving past it, but there is a little nagging detail here.

AKG said to me the following:

One more remark: You have A and B in R³, and C in the tangent space. It seems as though you should have them all in R³, or all in the tangent space.

And I replied as follows:

That is not consistent with any of the reading thus far. The rest of the chapter discussed forms defined on T_p\mathbb{R}^n that act on vectors in \mathbb{R}^n. Am I misunderstanding something?

On closer inspection of the text, it seems that I was wrong. But it seems as though there is actually a contradiction in the book. The notation &lt;\cdot , \cdot&gt; is explicitly said on p. 48 to denote vectors in a tangent space, and 1-forms on \mathbb{R}^n are said on p 50 to act on vectors of the form &lt;dx,dy&gt;, which means that they are vectors in the tangent space T_p\mathbb{R}^n. But looking at the diagram at the top of page 53, he plots the vectors V_1 and V_2 at the origin of a set of axes marked with x,y,z. This denotes the space \mathbb{R}^3, no? Well, if a 1-form acts on vectors from T_p\mathbb{R}^3, then I wonder why the axes aren't labeled dx,dy,dz?

OK, so here is what I'd like to know:

Just where do the vectors which are the arguments of a 1-form on T_p\mathbb{R}^n live? Do they live in the tangent space, or in \mathbb{R}^n itself?

And mathwonk: I'm not ignoring your geometric answer to Exercise 3.18. It's just that, as I said, it looks like we need an expression for the line l to move on to the other Exercises.

Boy I can't wait to be done with this chapter.

 

[mathwonk]

i pointed out long ago several of the many imprecisions and errors in this book, such as you are now noticing.

in this case there, is no big worry. i.e. there is a natural isomorphism between R^n and any of its tangent spaces. so there is no real problem in identifying one with the other.

of course it is no help in understanding the authors conventions.



according to mr bachman's earlier statements to me, the arguments for what he calls a 1 form are indeed elements of the tangent space.

 

[quetzalcoatl9]

The Flanders book, "Differential form and applications to the physical sciences" threw me for a loop with this. In the book, he started by referring to differential forms as "vectors". since the book came highly recommended to me, I began doubting everything that I thought I knew about forms until that point...

He was referring to them as vectors within the dual space (and later making a correspondence between them with vectors in E^3), which indeed they are...but he didn't lay that out until like 50 pages later, and it was unnecessarily confusing.

 

[mathwonk]

look, at every point of R^n, or any manifold, there is a tangent space. a form is a linear function on the tangent space, and a field of forms is a choice of a such a linear function on every tangent space. that's all there is to it. whatever language each person uses is only a distraction. just get the idea, then deal with each author's variations in language.


a confusion then is that for R^n the space itself is naturally isomorphic to every tangent space. so what?

if you prefer a book that actually writes down everything correctly and precisely the first time, read spivak instead of bachman.

 

[quantumdude]

in this case there, is no big worry. i.e. there is a natural isomorphism between R^n and any of its tangent spaces. so there is no real problem in identifying one with the other.

I understand that one space is a carbon copy of the other. If you recall, that was the reason I was moaning about the strange way in which he introduced the basis for T_p\mathbb{R}^2. But in this case there is a problem in identifying one with the other, because different two origins of the "home space" of V_1 and V_2 in Exercise 3.18 results in two different lines, and the line is the answer to the question. And the fact that that answer has to be used in the next 2 exercises makes it even worse.

To be honest, my advisees and I left this behind long ago just to move forward. We've finished all of chapters 4 and 5, and much of chapter 6 (up to Stokes' theorem). We just had no choice but to abandon this because of the deadline of the conference.

if you prefer a book that actually writes down everything correctly and precisely the first time, read spivak instead of bachman.

I would agree with that if you're talking about a course in advanced calculus. But I don't want to give up yet on the idea of a course in forms for college sophomores. But if I were going to suggest a course like that to my Department Chair, I can see now that I would want to invest the time putting it together myself, rather than just using this book.

OK, that's enough of that. I've solved Exercises 3.18 through 3.21. Solutions forthcoming shortly.

 

[quantumdude]

Chapter 3: Forms


Note: All symbols used in Exercises 3.18 through 3.21 have the same meaning.

Exercise 3.18
Let \omega=w_1dx \wedge dy +w_2dy \wedge dz +w_3dz \wedge dx.
Let V_1=&lt;a_1,a_2,a_3&gt; and V_2=&lt;b_1,b_2,b_3&gt; be vectors in T_p\mathbb{R}^3.
Let V_3=&lt;c_1,c_2,c_3&gt; be a vector in T_p\mathbb{R}^3 such that C=k_1V_1+k_2V_2. So the set {V_1,V_2,V_3} are dependent. That implies that det|V_3 V_1 V_2|=0.

Explicitly:

<br /> det [V_3 V_1 V_2]=\left |\begin{array}{ccc}c_1&amp;c_2&amp;c_3\\a_1&amp;a_2&amp;a_3\\b_1&amp;b_2&amp;b_3\end{array}\right|<br />

<br /> det [V_3 V_1 V_2]=c_1(a_2b_3-a_3b_2)-c_2(a_1b_3-a_3b_1)+c_3(a_1b_2-a_2b_1)<br />

Now let \omega act on V_1 and V_2. We obtain the following:

<br /> \omega (V_1,V_2)=w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)<br />

Upon comparing the expressions for det [V_3 V_2 V_1] and \omega (V_1,V_2) we find that \omega (V_1,V_2)=0 if w_1=c_3, w_2=c_1, and w_3=c_2. So the line l is the line that contains the vector V_3=&lt;w_2,w_3,w_1&gt;. So I can write down parametric equations for l as follows:

<br /> x=w_2t<br />
<br /> y=w_3t<br />
<br /> z=w_1t<br />
[/color]

 

[quantumdude]

Chapter 3: Forms


Exercise 3.19

Let ||V_1 \times V_2|| \equiv A, the area of the parallelogram spanned by V_1 and V_2.

Now look at \omega (V_1,V_2).

\omega (V_1,V_2)= w_1(a_1b_2-a_2b_1)+w_2(a_2b_3-a_3b_2)+w_3(a_3b_1-a_1b_3)

Recalling that V_3=&lt;w_2,w_3,w_1&gt; we have the following.

\omega (V_1,V_2)=V_3 \cdot (V_1 \times V_2)
\omega (V_2,V_2)=||V_3||A cos( \theta ),

where \theta is the angle between V_3 (and therefore l) and both V_1 and V_2. Noting that this dot product is maximized when \theta is 90 degrees, we have our result.

Exercise 3.20

Let N \equiv V_1 \times V_2.

Recalling the action of \omega on V_1 and V_2 from the last Exercise, we have the following.

\omega (V_1,V_2)=V_3 \cdot (V_1 \times V_2)

Noting the definition of N we see that we can immediately identify V_3 with V_{\omega}, and the desired result is obtained.

Exericise 3.21

Start by manipulating the expression given in the Exercise.

\omega= F_x dy \wedge dz - F_y dx \wedge dz + F_z dx \wedge dy
\omega = F_z dx \wedge dy + F_x dy \wedge dz - F_y dx \wedge dz
\omega = F_z dx \wedge dy + F_x dy \wedge dz + F_y dz \wedge dx

I used commutativity of 2-forms under addition to get to line 2, and anticommutativity of 1-forms under the wedge product to get to line 3.

Noting that V_3=&lt;c_1,c_2,c_3&gt;=&lt;w_2,w_3,w_1&gt; (Exercise 3.18) and noting that V_3=V_{\omega} (Exercise 3.20), it can be seen that V_{\omega}=&lt;F_x,F_y,F_z&gt;
[/color]

 

[mathwonk]

well on the positive side, some people actually learn more,by correcting errors of an imprecise book, than by plodding thriough one where all the i's are dotted for you. I think that may the case here. you seem to be learning a lot.

 

[quantumdude]

Too true. I sometimes hand out fallacious arguments to my students and ask them to find the errors.

Notes on Section 3.5 will be forthcoming shortly, and then we can finally get on to differential forms and integration.

Yahoo!

 

[phcatlantis]

Is it safe to say this thread is dead? I'm working through Bachman on my own and the discussion here has been pretty helpful.

 

[nrqed]

Calculation with differential forms

Hello folks,

I found a lovely little book online called A Geometric Approach to Differential Forms by David Bachman on the LANL arXiv. I've always wanted to learn this subject, and so I did something that would force me to: I've agreed to advise 2 students as they study it in preparation for a presentation at a local mathematics conference.

Since this was such a popular topic when lethe initially posted his Differential Forms tutorial, and since it is so difficult for me and my advisees to meet at mutually convenient times, I had a stroke of genius: Why not start a thread at PF?

Here is a link to the book:

http://xxx.lanl.gov/PS_cache/math/pdf/0306/0306194.pdf

As Bachman himself says, the first chapter is not necessary to learn the material, so I'd like to start with Chapter 2 (actually, we're at the end of Chapter 2, so hopefully I can stay 1 step ahead and lead the discussion!)

If anyone is interested, download the book and I'll post some of my notes tomorrow.

I ahve a question on the example of the integral presented in Example 3.3 (pages 40-41, from the hep archive).

He seems to go from dx^dy directly to dr^dt, where r and t are parametrizations of the upper half unit sphere, x= r cost, y = r sin t, z = sqrt(1- r^2), r ranging from 0 to 1 and t from 0 to 2 pi.

I don't understand that, it seems to me that dx^dy = r dr ^ dt.

Any one can help?

Thanks


Patrick

 

[George Jones]

The extra r is there.

(z^2) dx^dy was transformed to (1 - r^2) r dr^dt.

Regards,
George

 

[nrqed]

The extra r is there.

(z^2) dx^dy was transformed to (1 - r^2) r dr^dt.

Regards,
George

Yes, of course... Thanks

(I simply made the change of variables x,y -> r,t into dx^dy and got r dr^dt. Now I see that his \omega_{\phi(x,y)} calculates the Jacobian which is included automatically in the way I did it. Now I see that he literally meant to replace dx^dy by dr^dt without taking derivatives...that confused me).

Thanks..

On a related note...I know that I will sound stupid but I still find very confusing that the symbols "dx" and "dy" are used sometimes to represent infinitesimals and sometimes to represent differential forms.

Anyway...

 

[Doodle Bob]

On a related note...I know that I will sound stupid but I still find very confusing that the symbols "dx" and "dy" are used sometimes to represent infinitesimals and sometimes to represent differential forms.

Umm... that's on purpose since the one forms dx and dy are defined so that one can do the calculus without all this infinitesimal nonsense.

BTW whatevre is the obsession with infinitesimals? I thought that Bishop Berkley firmly nailed the last nail of their coffin way back in the 1600s. And Cauchy showed us how to do all of analysis and hence calculus without thinking once about them. Virtually no one that I know of in the research field actually thinks in terms of these. Don't we have enough non-computable numbers to deal with (e.g. the vast majority of irrational numbers) without willfully adding more?

 

[Hurkyl]

I thought that Bishop Berkley firmly nailed the last nail of their coffin way back in the 1600s.

I'm not sure what you mean, but I'm afraid you mean that using infinitessimals can make no sense! But we've had nonstandard analysis since the 1950s, which can be used to put infinitessimals on a perfectly rigorous foundation.

 

[George Jones]

I'm not sure what you mean, but I'm afraid you mean that using infinitessimals can make no sense! But we've had nonstandard analysis since the 1950s, which can be used to put infinitessimals on a perfectly rigorous foundation.

I'm not sure, but I think that Doodle Bob was referring to these when

Don't we have enough non-computable numbers to deal with (e.g. the vast majority of irrational numbers) without willfully adding more?

Regards,
George